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Ai lm giúp mk vs câu nào cũng được. Ai làm xong sớm nhất sẽ được tick
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Do \(x>0:\)
\(x^2+\dfrac{1}{x^2}=7\Leftrightarrow x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=9\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=9\Rightarrow x+\dfrac{1}{x}=3\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=3^3\Leftrightarrow x^3+3x.\dfrac{1}{x}.\left(x+\dfrac{1}{x}\right)+\dfrac{1}{x^3}=27\)
\(\Leftrightarrow x^3+3.1.3+\dfrac{1}{x^3}=27\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
\(\Rightarrow\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=7.18\Leftrightarrow x^5+\dfrac{1}{x}+x+\dfrac{1}{x^5}=126\)
\(\Leftrightarrow x^5+3+\dfrac{1}{x^5}=126\Rightarrow x^5+\dfrac{1}{x^5}=123\)
Ở dòng đầu gõ nhầm xíu \(\left(x+\dfrac{1}{x}\right)^2=9\) chứ ko phải \(\left(x+\dfrac{1}{x}\right)^3=9\)
a) (x + 5)2 - (x - 3)2 = 2x - 7
(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7
8(2x + 2)= 2x - 7
16x + 16 = 2x - 7
16x - 2x = - 7 - 16
14x = - 23
x = - 23/14
b) (2x - 3)(4x2 + 6x + 9) = 98
(2x)3 - 33 = 98
8x3 - 27 = 98
8x3 = 125
x3 = 125/8
x3 = (5/2)3
x = 5/2
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
Bài 2:
\(=\dfrac{-3x-1}{3\left(x-1\right)\left(x+1\right)}+\dfrac{5}{3\left(x-1\right)}+\dfrac{1}{3\left(x+1\right)}\)
\(=\dfrac{-3x-1+5x+5+x-1}{3\left(x-1\right)\left(x+1\right)}=\dfrac{3x+3}{3\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
b. Ta có: \(x+\dfrac{1}{x}=4\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3x^2.\dfrac{1}{x}+3x.\dfrac{1}{x^2}=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+12=64\)
\(\Rightarrow x^3+\dfrac{1}{x^3}=52\)
Lại có: \(x+\dfrac{1}{x}=4\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+2=16\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=14\)
Ta có: \(\left(x^3+\dfrac{1}{x^3}\right)\left(x^2+\dfrac{1}{x^2}\right)=52.14\)
\(\Rightarrow x^5+x+\dfrac{1}{x}+\dfrac{1}{x^5}=728\)
\(\Rightarrow x^5+\dfrac{1}{x^5}=724\)
a.
\(A=x^7+\dfrac{1}{x^7}\)
Ta có: \(\left(x^5+\dfrac{1}{x^5}\right)\left(x^2+\dfrac{1}{x^2}\right)=728.14\)
\(\Rightarrow x^7+x^3+\dfrac{1}{x^3}+\dfrac{1}{x^7}=10192\)
\(\Rightarrow x^7+\dfrac{1}{x^7}+52=10192\)
\(\Rightarrow x^7+\dfrac{1}{x^7}=10140\)
Ta có:
\(x^2+\dfrac{1}{x^2}=7\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2-2=7\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\)
\(\Rightarrow x+\dfrac{1}{x}=3\) ( Vì x > 0 )
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3.3=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}=18\)
Ta lại có:\(\left(x+\dfrac{1}{x}\right)\left(x^4+\dfrac{1}{x^4}\right)=x^5+x^3+\dfrac{1}{x^3}+\dfrac{1}{x^5}=x^5+\dfrac{1}{x^5}+18\)
Mặt khác:
\(\left(x+\dfrac{1}{x}\right)\left(x^4+\dfrac{1}{x^4}\right)=\left(x+\dfrac{1}{x}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)^2-2\right]\)
\(=\left(x+\dfrac{1}{x}\right)\left(7^2-2\right)\)
\(=3.47=141\)
\(\Rightarrow x^5+\dfrac{1}{x^5}+18=141\)
\(\Rightarrow x^5+\dfrac{1}{x^5}=123\)